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COMPUTER BASED TEST (CBT) Questions & Solutions
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PAGE # 1
7340010333
PART : PHYSICS (02-09-20)
Single Choice Type (,dy fodYih; izdkj)
This section contains 20 Single choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 ,dy fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. The magnetic field of a plane electromagnetic wave is
Ti)cty(200sin[103B 8
where c = 3 × 108 ms–1 is the speed of light.
the corresponding electric filed is :
(1) m/Vk)cty(200sin[9E
(2) m/Vk)cty(200sin[10E 6
(3) m/Vk)cty(200sin[103E 8
(4) m/Vk)cty(200sin[9E
Ans. (4)
Sol. E0 = cB0 C||BE
jik
E
= 3 × 108 × 3 × 10–8 )cty(200sin[ )k( = 9 )cty(200sin[ )k(
2. Pressure inside two soap bubbles are 1.01 and 1.02 atmosphere, respectively. The ratio of their
volumes is :
(1) 0.8 : 1 (2) 4 : 1 (3) 8 : 1 (4) 2 : 1
Ans. (3)
Sol. P1 – P0 = 1RT4
P2 – P0 = 2RT4
Dividing,
1
2
RR
21
R1 = 2R2
18
RR8
RR
VV
32
32
32
31
2
1
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3. A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of
the earth's radius Re. By firing rockets attached to it, its speed is instantaneously increased in the
direction of its motion so that it becomes 23 times larger. Due to this the farthest distance from the
centre of the earth that the satellite reaches is R. Value of R is :
(1) 2Re (2) 3Re (3) 2.5Re (4) 4Re
Ans. (2) Sol.
eRMGV
From energy conversation
2min
max
2
emV
21
RGMmV
23m
21
RGMm
….(i)
From angular momentum conversation
maxmine RVVR23
….(ii)
Eliminating Vmin from equation (i) and (ii) we get
Rmax = 3Re
4. In a radioactive material, fraction of active material remaining after tie t is 169 . The fraction that was
remaining after t/2 is :
(1) 53 (2)
43 (3)
87 (4)
54
Ans. (2)
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PAGE # 3
7340010333
Sol. N = No e–t …(1)
N' = N0 2t
e
….(2)
from (1) & (2)
43
169
NN
N'N 2
121
00
5. In the circuit shown in the figure, the total charge is 750 C and the voltage across capacitor C2 is 20 V.
Then the charge on capacitor C2 is :
(1) 160 C (2) 450 C (3) 590 C (4) 650 C
Ans. (3) Sol.
V1q3
q2
Q
C2
C1 = 15F
C3 = 8 F
V2 = 20V
q2 + q3 = 750
q2 + 160 = 750
q2 = 590
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PAGE # 4
7340010333
6.
Consider a gas of triatomic molecules. The molecules rigid rods whose vertices are occupied by atoms.
The internal energy of a mole of the gas at temperature T is :
(1) RT29 (2) RT
23 (3) 3RT (4) RT
25
Ans. (3)
Sol. RT26RT
2fU = 3RT
7. Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is
measured. it should correctly be recorded as :
(1) 2.124 cm (2) 2.123 cm (3) 2.125 cm (4) 2.121 cm
Ans. (1) Sol. Thickness = M.S. Reading + Circular Scale Reading (L.C.)
Here, LC = 50
1.0 = 0.002 cm per division
8. In Young's double slit experiment, light of 500 nm is used to produce and interference pattern. When
the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the
distance screen is close to :
(1) 0.17º (2) 0.07º (3) 0.57º (4) 1.7º
Ans. (3)
Sol. 0 = 5
9
10510500
d
= 10–2 Radian = 0.57°
9. When a diode is forward biased, it has a voltage drop of 0.5 v. the safe limit of current through the diode
is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be
connected in series with the diode so that the current does not exceed the safe limit is :
(1) 50 (2) 300 (3) 200 (4) 100
Ans. (4)
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PAGE # 5
7340010333
Sol.
1.5 V
R 10–2A 0.5 volt
Vdiode = 0.5 volt
VR = 1.5 – 0.5 = 1 volt
iR = 1
R = i1 = 210
1 = 100
10. An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in a
magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency ,
the average power loss in the lop due to Joule heating is :
(1) R2Bba 22222 (2)
RBba 22222 (3)
RabB (4) zero
Ans. (1)
Sol. = NAB cost
<P> = <R
2 >
=
R
tcosBA 2222
=
21
RBA 222
= R2
Bba 2222 (2)
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PAGE # 6
7340010333
11. Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm, 50 turns
and carrying current (Ampere) in units of 0 is :
(1) 3250 (2) 35 (3) 3500 (4) 350
Ans. (3)
Sol. B = 0i50 6104 cos30º
100
[sin 30° + sin 30°]
2 × 75 × 10
21
21
3i0
i3500i
31500 00
3500 Ans.
12. A charged particle carrying charge 1 C is moving with velocity k4j3i2 ms–1. If an external
magnetic field of k6j3i5 × 10–3T exists in the region where the particle is moving then the force on
the particle is N10F 9
. the vector F
is :
(1) k9j32i30 (2) k9.0j32i0.3
(3) k09.0j32.0i30.0 (4) k90j320i300
Ans. (1)
Sol.
635432kji
1010F 36
= ( k9j32i30 ) × 10–9 N
13. Two isolated conducting spheres S1 and S2 of radius R32 and R
31 have 12 C and –3C charges,
respectively, and are at a large distance from each other, They are now connected by a conducting
wire. A long time after this is done the charges on S1 and S2 are respectively :
(1) 6 C and 3 C (2) 4.5 C of both
(3) +4.5 C and –4.5 C (4) 3 C and 6 C
Ans. (1)
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PAGE # 7
7340010333
Sol. Total charge = 12 – 3 = 9C
If final charges are q1 and q2
12
RR
2
1
2
1
q1 = 6C
q2 = 3C
14. A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and collides
with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result
of the collision making angle before momentarily coming to rest. if the rod has mass M = 2 kg, and
length = 1m, the value of is approximately (take g = 10 m/s2)
(1) 69º (2) 63º (3) 55º (4) 49º
Ans. (2) Sol. Angular momentum
3m2mmv
22
mv= 2m35
= 5v3
)cos1(mg)cos1(2
mg221 2
2
22
25v9m
35
21
= 2mg ( 1–cos)
2mv25
3
= 2mg (1–cos)
cos1102
36103
1– 5027 = cos
cos = 5023
= 63°
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PAGE # 8
7340010333
15. A balloon filled with helium (32º C and 1.7 atm) bursts. Immediately afterwards the expansion of helium
can be considered as :
(1) reversible adiabatic (2) reversible isothermal
(3) irreversible isothermal (4) irreversible adiabatic
Ans. (4) Sol. Theory based
16. When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum
kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close
to : (1) 1.02 eV (2) 0.61 eV (3) 0.52 eV (4) 0.81 eV
Ans. (2)
Sol. KEmax =
hc = 500hc ……(i)
Now, 3KEmax = 200hc ……(ii)
From equation (i) and (ii)
)i()ii( =
500hc
200hc
13
Put the value of hc = 1237.5 and solving = 0.61 eV
17. Model a torch battery of length to be made up of a thin cylindrical bar of radius 'a' and a concentric
thin cylindrical shell of radius 'b' filled in between with an electrolyte of resistivity (see figure). If the
battery is connected to a resistance of value R, the maximum Joule heating in R will take place for :
(1)
ab
2R
(2)
abnR
(3)
abn2R
(4)
abn
2R
Ans. (4)
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PAGE # 9
7340010333
Sol.
ab
The resistance of small element
R = r2
dr
R =
b
ardr
2
R = abn
2
18. Moment of inertia of a cylinder of mass m> length L and radius R about an axis passing through its
centre and perpendicular to the axis of the cylinder is = M
12L
4R 22
. If such a cylinder is to be made
for a given mass of a material, the ratio RL for it to have minimum possible is :
(1) 32 (2)
32 (3)
23 (4)
23
Ans. (4) Sol. Let a cylinder of mass m1 length L and sodium R then Let take elementary disc of sodium R and
trickiness dx at a distance of x from axis Oo' then moment of inertia about Oo' as this element
dx
R
22
dmx4
dmRdI
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PAGE # 10
7340010333
I = 22/Ln
2/Ln
2xdx
LM
4dmRdI
12
ML4
MRI22
I = 12
MLL
V4M 2
I = 12
MLL4
MV 2
012
L2ML4
mVdLdI
2
V = 3L32 32 L
32LR
23
RL
19. A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of
mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is produced at
the lower and the rope. What is the wavelength of the wavetrain (in cm) when it reaches the top of the
rope ?
(1) 12 (2) 6 (3) 9 (4) 3
Ans. (1)
Sol. V = f
2
2
1
1 VV
2 = 11
2
VV
= 11
2
TT
T2 = 8g (Top)
1g2g8 T1 = 2g (Bottom)
= 21 = 12 cm
20. A 750 Hz, 20 V (rms) source is connected to a resistance of 100 , an inductance of 0.1803 H and a
capacitance of 10F all in series. the time in which the resistance (heat capacity CºJ2 ) will get heated by
10ºc. (assume no loss of heat to the surroundings) is close to :
(1) 365 s (2) 418 s (3) 348 s (4) 245 s
Ans. (3)
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PAGE # 11
7340010333
Sol.
R = 100, XL = L = 0.1803 × 750 × 2 = 850 , XC = 23.21750210
1C1
5
2CL
2 )XX(RZ
= 22 )23.21850(100 = 834.77
H = t)ms(Rti2rms
10)2(t10083520
83520
t = 348.61 sec
21. A Bakelite beaker has volume capacity of 500 cc at 30ºC. When it is partially filled with Vm volume (at
30ºC) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is
varied. If (beaker) = 6 × 10–6 ºC–1, where is the coefficient of volume expansion, then Vm (in cc) is close
to ………….
Ans. 20.00
Sol. Vm = VC
VmM T = VCC T
Vm = m
CCV = 4
6
105.1106500
= 20 cc
22. A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its
axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the
platform. What will be the rotational speed (in rpm) of the platform when the person reaches its
centre….
Ans. 09.00
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PAGE # 12
7340010333
Sol. '2
MR2
MRmR22
2
'2
20052
20080
' = 9 rev/sec.
23. When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of
15 cm within it. If the contact angle between the liquid and glass to close to 0º, the surface tension of
the liquid, in milliNewton m–1, is [(liquid) = 900 kgm–3, g = 10 ms–2] (Given answer in closed integer)
Ans. 101
Sol. ghrT2
T = 2
grh = 1012
1090010151015 25
milliNewton m–1
24. An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of 15
cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up
to a height of 30 cm the same observer can see the edge at the bottom of the jar. If the refractive index
of the liquid is 100N , where N is an integer, the value of N is
Ans. 158
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PAGE # 13
7340010333
Sol.
P15 cm
30 cm
30 cm
45 cm
r
i
15 cm
15 cm45º
From Shell's Law
×sini = 1 × sinr
× 22 3015
15
= 1 × sin 45º
= 25 = 158 × 10–2
25. A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum
height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the
ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching
the ball, the value of F (in N) is (g = 10 ms–2)
Ans. 150.00
Sol. From work energy theorem F(0.2) – mg(20) = 0
F = 2.0)20(mg
=
2
200mg
= 2
2001015.0
= 150.00 N
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